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I find in general that the pull-back of a section of a vector bundle is a section of the pull-back bundle, but this seems to be false for the cotangent bundle.

Let $\phi:M\to N$ be a smooth map, $\omega\in \Gamma(T^*N)$ be a $1$-form and $X\in \Gamma(TM)$ be a vector field.

Then, since $(\phi^*\omega)(X)=\omega(\phi_*X)$, we can say that $\phi^*\omega$ is a $1$-form of $T^*M\ncong\phi^*T^*N$, in particular the fibers of $T^*N$ and $T^*M$ are different and thus the pull-back bundle $\phi^*T^*N$ cannot be equivalent to $T^*M$.

$\textbf{My question}$: why?

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  • $\begingroup$ Are you asking whether the pullback of a $T^*N$ should be $T^*M$? If so this is not the case; if not could you clarify your question? $\endgroup$
    – Ben
    Mar 6, 2019 at 14:48
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    $\begingroup$ Also a pull back of a section is again section, as there is a map $\Gamma(T^*N) \to \Gamma(T^*M)$, but this is neither injective nor surjective, and not all sections are pulled back from $N$ $\endgroup$
    – Ben
    Mar 6, 2019 at 14:51
  • $\begingroup$ My question is: why in general the pull back of a section is a section in the pull-back bundle (like here math.stackexchange.com/questions/245965/…), whereas for the cotangent bundle this is not true? $\endgroup$
    – Bellem
    Mar 6, 2019 at 14:56

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You are confused by two different notions of pullback. The pullback of $f:M\rightarrow N$ by $g:P\rightarrow N$ is $(x,y)\in M\times P$ such that $f(x)=g(x)$.

A $1$-form defined on $N$ is a morphism $\omega:N\rightarrow T^*N$, if you have $\phi:M\rightarrow N$, you cannot defined the pullback of $\omega$ in the previous sense since the target of $\phi$ is not $T^*N$, there is another notion of pullback of a $1$-form derfined on $M$ by $(\phi^*\omega)_x(u)=\omega(d\phi_x(u))$.

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  • $\begingroup$ Ok. But in general I can pull-back a section by a continuous map and that is a section in the pull-back bundle, like here: math.stackexchange.com/questions/245965/…. I don't understand why this case should be different... $\endgroup$
    – Bellem
    Mar 6, 2019 at 15:00
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    $\begingroup$ You can call two different notions with the same name. $\endgroup$ Mar 6, 2019 at 15:04
  • $\begingroup$ Ok I totally agree. My point is that a $1$-form is a section of a vector bundle, so I can pull-back it in the sense of pulling back a section of a vector bundle to a section of the pull-back bundle. I don't understand why if I do this for a general section (not a differential form) in a general vector bundle that works, whereas for a differential form that does not work. I.e. the pull-back section is not a section of the pull-back vector bundle... $\endgroup$
    – Bellem
    Mar 6, 2019 at 15:08
  • $\begingroup$ You can do that for forms too, but as you remarked $\phi^*TN$ is not $T^*M$, so by using that definition of pullback you obtain an element of $\phi^*TN$ which is not a form defined on $M$ but a section of $\phi^*T^*N$. $\endgroup$ Mar 6, 2019 at 15:10
  • $\begingroup$ Ok, I read your modified comment and now it is clear to me, thank you! $\endgroup$
    – Bellem
    Mar 6, 2019 at 19:46

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